In this section, we present our results.
Theorem 2.1
Let
T
be a time scale. For
\(\diamondsuit_{\alpha}\)
differentiable
\(f: [ 0,h ] \cap T \to R\), with
\(f ( 0 ) = 0\)
we have
$$ \int _{0}^{h} \bigl\vert {f^{k} } \bigr\vert ^{\diamondsuit_{\alpha}} ( t )\diamondsuit_{\alpha}t \le h^{k - 1} \int _{0}^{h} \bigl\vert {f^{\diamondsuit_{\alpha}} } \bigr\vert ^{k} ( t )\diamondsuit_{\alpha}t. $$
(2)
Proof
Starting with the left side of (2), we obtain
$$\begin{aligned} \int _{0}^{h} \bigl\vert f^{k} \bigr\vert ^{\diamondsuit_{\alpha}} ( t )\diamondsuit _{\alpha}( t ) =& \int _{0}^{h} \bigl\vert f \cdot f^{k - 1} \bigr\vert ^{\diamondsuit_{\alpha}} ( t )\diamondsuit_{\alpha}( t ) \\ =& \int _{0}^{h} \bigl\vert f^{k - 1} f^{\diamondsuit_{\alpha}} + \alpha f^{\sigma}\bigl( f^{k - 1} \bigr)^{\Delta}+ ( 1 - \alpha )f^{\rho}\bigl( f^{k - 1} \bigr)^{\nabla}\bigr\vert ( t )\diamondsuit_{\alpha}( t ) \\ =& \alpha \int _{0}^{h} \bigl\vert f^{k - 1} f^{\diamondsuit _{\alpha}} + \alpha f^{\sigma}\bigl( f^{k - 1} \bigr)^{\Delta}+ ( 1 - \alpha )f^{\rho}\bigl( f^{k - 1} \bigr)^{\nabla}\bigr\vert ( t )\Delta t \\ &{}+ ( 1 - \alpha ) \int _{0}^{h} \bigl\vert f^{k - 1} f^{\diamondsuit_{\alpha}} + \alpha f^{\sigma}\bigl( f^{k - 1} \bigr)^{\Delta}+ ( 1 - \alpha )f^{\rho}\bigl( f^{k - 1} \bigr)^{\nabla}\bigr\vert ( t )\nabla t \\ \le&\alpha \int _{0}^{h} \bigl\vert f^{k - 1} f^{\diamondsuit _{\alpha}} \bigr\vert ( t )\Delta t + \alpha^{2} \int _{0}^{h} \bigl\vert f^{\sigma}\bigl( f^{k - 1} \bigr)^{\Delta}\bigr\vert ( t )\Delta t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \bigl\vert f^{\rho}\bigl( f^{k - 1} \bigr)^{\nabla}\bigr\vert ( t )\Delta t + ( 1 - \alpha ) \int _{0}^{h} \bigl\vert f^{k - 1} f^{\diamondsuit_{\alpha}} \bigr\vert ( t )\nabla t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \bigl\vert f^{\sigma}\bigl( f^{k - 1} \bigr)^{\Delta}\bigr\vert ( t )\nabla t + ( 1 - \alpha )^{2} \int _{0}^{h} \bigl\vert f^{\rho}\bigl( f^{k - 1} \bigr)^{\nabla}\bigr\vert ( t )\nabla t. \end{aligned}$$
Using Definition 1.3, we get
$$\begin{aligned} \int _{0}^{h} \bigl\vert f^{k} \bigr\vert ^{\diamondsuit_{\alpha}} ( t )\diamondsuit _{\alpha}( t ) \le&\alpha \int _{0}^{h} \bigl\vert \alpha f^{k - 1} f^{\Delta}+ ( 1 - \alpha )f^{k - 1} f^{\nabla}\bigr\vert ( t )\Delta t \\ & {}+ \alpha^{2} \int _{0}^{h} \bigl\vert f^{\sigma}\bigl( f^{k - 1} \bigr)^{\Delta}\bigr\vert ( t )\Delta t + \alpha ( 1 - \alpha ) \int _{0}^{h} \bigl\vert f^{\rho}\bigl( f^{k - 1} \bigr)^{\nabla}\bigr\vert ( t )\Delta t \\ & {}+ ( 1 - \alpha ) \int _{0}^{h} \bigl\vert \alpha f^{k - 1} f^{\Delta}+ ( 1 - \alpha )f^{k - 1} f^{\nabla}\bigr\vert ( t )\nabla t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \bigl\vert f^{\sigma}\bigl( f^{k - 1} \bigr)^{\Delta}\bigr\vert ( t )\nabla t + ( 1 - \alpha )^{2} \int _{0}^{h} \bigl\vert f^{\rho}\bigl( f^{k - 1} \bigr)^{\nabla}\bigr\vert ( t )\nabla t \\ \le&\alpha^{2} \int _{0}^{h} \bigl\vert f^{k - 1} f^{\Delta}\bigr\vert ( t )\Delta t + \alpha ( 1 - \alpha ) \int _{0}^{h} \bigl\vert f^{k - 1} f^{\nabla}\bigr\vert ( t )\Delta t \\ & {}+ \alpha^{2} \int _{0}^{h} \bigl\vert f^{\sigma}\bigl( f^{k - 1} \bigr)^{\Delta}\bigr\vert ( t )\Delta t + \alpha ( 1 - \alpha ) \int _{0}^{h} \bigl\vert f^{\rho}\bigl( f^{k - 1} \bigr)^{\nabla}\bigr\vert ( t )\Delta t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \bigl\vert f^{k - 1} f^{\Delta}\bigr\vert ( t )\nabla t + ( 1 - \alpha )^{2} \int _{0}^{h} \bigl\vert f^{k - 1} f^{\nabla}\bigr\vert ( t )\nabla t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \bigl\vert f^{\sigma}\bigl( f^{k - 1} \bigr)^{\Delta}\bigr\vert ( t )\nabla t + ( 1 - \alpha )^{2} \int _{0}^{h} \bigl\vert f^{\rho}\bigl( f^{k - 1} \bigr)^{\nabla}\bigr\vert ( t )\nabla t. \end{aligned}$$
We find that
$$\begin{aligned} \int _{0}^{h} \bigl\vert f^{\sigma}\bigl( f^{k - 1} \bigr)^{\Delta}\bigr\vert ( t )\Delta t =& \int _{0}^{h} \bigl\vert f^{\sigma}\bigl( f \cdot f^{k - 2} \bigr)^{\Delta}\bigr\vert ( t ) \Delta t \\ =& \int _{0}^{h} \bigl\vert f^{\sigma}\bigl( f^{\Delta}f^{k - 2} + f^{\sigma}\bigl( f \cdot f^{k - 3} \bigr)^{\Delta}\bigr) \bigr\vert ( t )\Delta t \\ & {}\vdots \\ =& \int _{0}^{h} \bigl\vert f^{\sigma}\bigl( f^{\Delta}f^{k - 2} + f^{\sigma}f^{\Delta}f^{k - 3} + \cdots+ \bigl( f^{\sigma}\bigr)^{2} f^{\Delta}\bigr) \bigr\vert ( t )\Delta t \\ =& \int _{0}^{h} \bigl\vert f^{\sigma}f^{k - 2} + \bigl( f^{\sigma}\bigr)^{2} f^{k - 3} + \cdots+ \bigl( f^{\sigma}\bigr) \bigr\vert \bigl\vert f^{\Delta}\bigr\vert ( t )\Delta t \\ =& \int _{0}^{h} \Biggl\vert \sum _{n = 0}^{k - 2} f^{n} \bigl( f^{\sigma}\bigr)^{k - 1 - n} \Biggr\vert \bigl\vert f^{\Delta}\bigr\vert ( t )\Delta t. \end{aligned}$$
Similarly,
$$\int _{0}^{h} \bigl\vert f^{\rho}\bigl( f^{k - 1} \bigr)^{\nabla}\bigr\vert ( t )\Delta t = \int _{0}^{h} \Biggl\vert \sum _{n = 0}^{k - 2} f^{n} \bigl( f^{\rho}\bigr)^{k - 1 - n} \Biggr\vert \bigl\vert f^{\nabla}\bigr\vert ( t )\nabla t. $$
Therefore,
$$\begin{aligned} \int _{0}^{h} \bigl\vert f^{k} \bigr\vert ^{\diamondsuit _{\alpha}} ( t )\diamondsuit_{\alpha}t \le& \alpha^{2} \int _{0}^{h} \bigl\vert f^{k - 1} f^{\Delta}\bigr\vert ( t )\Delta t + \alpha ( 1 - \alpha ) \int _{0}^{h} \bigl\vert f^{k - 1} f^{\nabla}\bigr\vert ( t )\Delta t \\ & {}+ \alpha^{2} \int _{0}^{h} \Biggl\vert \sum _{n = 0}^{k - 2} f^{n} \bigl( f^{\sigma}\bigr)^{k - 1 - n} \Biggr\vert \bigl\vert f^{\Delta}\bigr\vert ( t )\Delta t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \Biggl\vert \sum _{n = 0}^{k - 2} f^{n} \bigl( f^{\rho}\bigr)^{k - 1 - n} \Biggr\vert \bigl\vert f^{\nabla}\bigr\vert ( t )\Delta t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \bigl\vert f^{k - 1} f^{\Delta}\bigr\vert ( t )\nabla t + ( 1 - \alpha )^{2} \int _{0}^{h} \bigl\vert f^{k - 1} f^{\nabla}\bigr\vert ( t )\nabla t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \Biggl\vert \sum _{n = 0}^{k - 2} f^{n} \bigl( f^{\sigma}\bigr)^{k - 1 - n} \Biggr\vert \bigl\vert f^{\Delta}\bigr\vert \nabla t \\ & {}+ ( 1 - \alpha )^{2} \int _{0}^{h} \Biggl\vert \sum _{n = 0}^{k - 2} f^{n} \bigl( f^{\rho}\bigr)^{k - 1 - n} \Biggr\vert \bigl\vert f^{\nabla}\bigr\vert ( t )\nabla t \\ = &\alpha^{2} \int _{0}^{h} \Biggl( \bigl\vert f^{k - 1} \bigr\vert + \Biggl\vert \sum_{n = 0}^{k - 2} f^{n} \bigl( f^{\sigma}\bigr)^{k - 1 - n} \Biggr\vert \Biggr) \bigl\vert f^{\Delta}( t ) \bigr\vert \Delta t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \Biggl( \bigl\vert f^{k - 1} \bigr\vert + \Biggl\vert \sum_{n = 0}^{k - 2} f^{n} \bigl( f^{\rho}\bigr)^{k - 1 - n} \Biggr\vert \Biggr) \bigl\vert f^{\nabla}( t ) \bigr\vert \Delta t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \Biggl( \bigl\vert f^{k - 1} \bigr\vert + \Biggl\vert \sum_{n = 0}^{k - 2} f^{n} \bigl( f^{\sigma}\bigr)^{k - 1 - n} \Biggr\vert \Biggr) \bigl\vert f^{\Delta}( t ) \bigr\vert \nabla t \\ & {}+ ( 1 - \alpha )^{2} \int _{0}^{h} \Biggl( \bigl\vert f^{k - 1} \bigr\vert + \Biggl\vert \sum_{n = 0}^{k - 2} f^{n} \bigl( f^{\rho}\bigr)^{k - 1 - n} \Biggr\vert \Biggr) \bigl\vert f^{\nabla}( t ) \bigr\vert \nabla t \\ \le&\alpha^{2} \int _{0}^{h} \Biggl( \bigl\vert f^{k - 1} \bigr\vert + \sum_{n = 0}^{k - 2} \bigl\vert f^{n} \bigl( f^{\sigma}\bigr)^{k - 1 - n} \bigr\vert \Biggr) \bigl\vert f^{\Delta}( t ) \bigr\vert \Delta t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \Biggl( \bigl\vert f^{k - 1} \bigr\vert + \sum_{n = 0}^{k - 2} \bigl\vert f^{n} \bigl( f^{\rho}\bigr)^{k - 1 - n} \bigr\vert \Biggr) \bigl\vert f^{\nabla}( t ) \bigr\vert \Delta t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \Biggl( \bigl\vert f^{k - 1} \bigr\vert + \sum_{n = 0}^{k - 2} \bigl\vert f^{n} \bigl( f^{\sigma}\bigr)^{k - 1 - n} \bigr\vert \Biggr) \bigl\vert f^{\Delta}( t ) \bigr\vert \nabla t \\ & {}+ ( 1 - \alpha )^{2} \int _{0}^{h} \Biggl( \bigl\vert f^{k - 1} \bigr\vert + \sum_{n = 0}^{k - 2} \bigl\vert f^{n} \bigl( f^{\rho}\bigr)^{k - 1 - n} \bigr\vert \Biggr) \bigl\vert f^{\nabla}( t ) \bigr\vert \nabla t \\ = &\alpha^{2} \int _{0}^{h} \Biggl( \sum _{n = 0}^{k - 1} \bigl\vert f^{n} \bigr\vert \bigl\vert \bigl( f^{\sigma}\bigr)^{k - 1 - n} \bigr\vert \Biggr) \bigl\vert f^{\Delta}( t ) \bigr\vert \Delta t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \Biggl( \sum _{n = 0}^{k - 1} \bigl\vert f^{n} \bigr\vert \bigl\vert \bigl( f^{\rho}\bigr)^{k - 1 - n} \bigr\vert \Biggr) \bigl\vert f^{\nabla}( t ) \bigr\vert \Delta t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \Biggl( \sum _{n = 0}^{k - 1} \bigl\vert f^{n} \bigr\vert \bigl\vert \bigl( f^{\sigma}\bigr)^{k - 1 - n} \bigr\vert \Biggr) \bigl\vert f^{\Delta}( t ) \bigr\vert \nabla t \\ & {}+ ( 1 - \alpha )^{2} \int _{0}^{h} \Biggl( \sum _{n = 0}^{k - 1} \bigl\vert f^{n} \bigr\vert \bigl\vert \bigl( f^{\rho}\bigr)^{k - 1 - n} \bigr\vert \Biggr) \bigl\vert f^{\nabla}( t ) \bigr\vert \nabla t. \end{aligned}$$
Consider \(g ( t ) = \int _{0}^{t} \vert f^{\diamondsuit_{\alpha}} ( s ) \vert \diamondsuit _{\alpha}s\). Then we have \(g^{\Delta}( t ) = \vert f^{\Delta}( t ) \vert \), \(g^{\nabla}( t ) = \vert f^{\nabla}( t ) \vert \), and \(\vert f \vert \le g\), so that \(g ( t ) = \int _{0}^{t} \vert f^{\diamondsuit_{\alpha}} ( s ) \vert \diamondsuit_{\alpha}s \ge \vert \int _{0}^{t} f^{\diamondsuit_{\alpha}} ( s )\diamondsuit_{\alpha}s \vert = \vert f ( t ) - f ( 0 ) \vert = \vert f ( t ) \vert \).
The above inequality becomes
$$\begin{aligned} \int _{0}^{h} \bigl\vert f^{k} \bigr\vert ^{\diamondsuit_{\alpha}} ( t )\diamondsuit _{\alpha} \le& \alpha^{2} \int _{0}^{h} \Biggl( \sum _{n = 0}^{k - 1} g^{n} \bigl( g^{\sigma}\bigr)^{k - 1 - n} \Biggr) \bigl( g^{\Delta}\bigr) ( t )\Delta t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \Biggl( \sum _{n = 0}^{k - 1} g^{n} \bigl( g^{\rho}\bigr)^{k - 1 - n} \Biggr) \bigl( g^{\nabla}\bigr) ( t )\Delta t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \Biggl( \sum _{n = 0}^{k - 1} g^{n} \bigl( g^{\sigma}\bigr)^{k - 1 - n} \Biggr) \bigl( g^{\Delta}\bigr) ( t )\nabla t \\ & {}+ ( 1 - \alpha )^{2} \int _{0}^{h} \Biggl( \sum _{n = 0}^{k - 1} g^{n} \bigl( g^{\rho}\bigr)^{k - 1 - n} \Biggr) \bigl( g^{\nabla}\bigr) ( t )\nabla t \\ = &\alpha^{2} \int _{0}^{h} \bigl( g^{k} \bigr)^{\Delta}( t )\Delta t + \alpha ( 1 - \alpha ) \int _{0}^{h} \bigl( g^{k} \bigr)^{\nabla}( t )\Delta t \\ & {}+ \alpha ( 1 - \alpha ) \int _{0}^{h} \bigl( g^{k} \bigr)^{\Delta}( t )\nabla t + ( 1 - \alpha )^{2} \int _{0}^{h} \bigl( g^{k} \bigr)^{\nabla}( t )\nabla t \\ = &\alpha \biggl[ \alpha \int _{0}^{h} \bigl( g^{k} \bigr)^{\Delta}\Delta t + ( 1 - \alpha ) \int _{0}^{h} \bigl( g^{k} \bigr)^{\Delta}\nabla t \biggr] \\ & {}+ ( 1 - \alpha ) \biggl[ \int _{0}^{h} \alpha \bigl( g^{k} \bigr)^{\nabla}\Delta t + ( 1 - \alpha ) \int _{0}^{h} \bigl( g^{k} \bigr)^{\nabla}\nabla t \biggr] \\ = &\alpha \int _{0}^{h} \bigl( g^{k} \bigr)^{\Delta}\diamondsuit_{\alpha}+ ( 1 - \alpha ) \int _{0}^{h} \bigl( g^{k} \bigr)^{\nabla}\diamondsuit _{\alpha}= \int _{0}^{h} \bigl( g^{k} \bigr) ( t )^{\diamondsuit_{\alpha}} \diamondsuit_{\alpha}\\ =& g^{k} ( t ) | _{0}^{h} = g^{k} ( h ) - g^{k} ( 0 ) = \bigl[ g ( h ) \bigr]^{k} = \biggl[ \int _{0}^{h} \bigl\vert f^{\diamondsuit _{\alpha}} ( s ) \bigr\vert \diamondsuit_{\alpha}s \biggr]^{k}. \end{aligned}$$
By using Hölder’s inequality with indices \(p = \frac{k}{k - 1}\) and \(q = k\), we obtain
$$\begin{aligned} \biggl[ \int _{0}^{h} 1 \cdot \bigl\vert {f^{\diamondsuit_{\alpha}} ( s )} \bigr\vert \diamondsuit_{\alpha}s \biggr]^{k} \le& \biggl[ \biggl( \int _{0}^{h} 1^{\frac{k}{k - 1}} \diamondsuit _{\alpha}s \biggr)^{\frac{k - 1}{k}} \biggl( \int _{0}^{h} \bigl\vert {f^{\diamondsuit_{\alpha}} ( s )} \bigr\vert ^{k} \diamondsuit_{\alpha}s \biggr)^{\frac{1}{k}} \biggr]^{k} \\ = & \biggl( \int _{0}^{h} \diamondsuit_{\alpha}s \biggr)^{k - 1} \biggl( \int _{0}^{h} \bigl\vert {f^{\diamondsuit _{\alpha}} ( s )} \bigr\vert ^{k} \diamondsuit_{\alpha}s \biggr) \\ = & \bigl( s |_{0}^{h} \bigr)^{k - 1} \int _{0}^{h} \bigl\vert {f^{\diamondsuit_{\alpha}} ( s )} \bigr\vert ^{k} \diamondsuit_{\alpha}s \\ = &h^{k - 1} \int _{0}^{h} \bigl\vert {f^{\diamondsuit_{\alpha}} ( s )} \bigr\vert ^{k} \diamondsuit_{\alpha}s, \end{aligned}$$
hence the proof is complete. □
Theorem 2.2
Let
\(\omega ( t )\)
be positive and continuous on
\(( 0,h )\), with
\(\int _{0}^{h} \omega^{1 - q} ( t )\Delta t < \infty\), \(q > 1\). For differentiable
\(f: [ 0,h ] \to\mathbb{R}\)
with
\(f ( 0 ) = 0\)
we have
$$ \int _{0}^{h} \bigl\vert {f^{k} } \bigr\vert ^{\Delta}\Delta t \le \biggl( \int _{0}^{h} \omega^{1 - q} \Delta t \biggr)^{\frac{k}{q}} \biggl( \int _{0}^{h} \omega \bigl\vert {f^{\Delta}} \bigr\vert ^{p} \Delta t \biggr)^{\frac{k}{q}}, $$
(3)
where
\(p > 1\)
and
\(\frac{1}{p} + \frac{1}{q} = 1\).
Proof
We take \(g ( t ) = \int _{0}^{t} \vert f^{\Delta}( s ) \vert \Delta s\). Then \(\vert f ( t ) \vert \le g ( t )\), \(g^{\Delta}( t ) = \vert f^{\Delta}( t ) \vert \), so we have
$$\begin{aligned} \int _{0}^{h} \bigl\vert {f^{k} } \bigr\vert ^{\Delta}\Delta t =& \int _{0}^{h} \Biggl\vert {\sum _{k = 0}^{n - 1} f^{k} \bigl( f^{\sigma}\bigr)^{n - 1 - k} } \Biggr\vert \bigl\vert {f^{\Delta}} \bigr\vert ( t )\Delta t \\ \le& \int _{0}^{h} \Biggl( \sum _{k = 0}^{n - 1} g^{k} \bigl( g^{\sigma}\bigr)^{n - 1 - k} \Biggr) \bigl( g^{\Delta}\bigr) ( t )\Delta t = \int _{0}^{h} \bigl( g^{k} \bigr)^{\Delta}\Delta t \\ =& g^{k} ( h ) - g^{k} ( 0 ) = g^{k} ( h ) = \biggl( \int _{0}^{h} \bigl\vert {f^{\Delta}} \bigr\vert ( t )\Delta t \biggr)^{k} \\ =& \biggl( \int _{0}^{h} \omega^{ - \frac{1}{p}} \omega ^{\frac{1}{p}} \bigl\vert {f^{\Delta}} \bigr\vert ( t )\Delta t \biggr)^{k} \\ \le& \biggl[ \biggl( \int _{0}^{h} \bigl( \omega^{ - \frac{1}{p}} \bigr)^{q} \Delta t \biggr)^{\frac{1}{q}} \biggl( \int _{0}^{h} \bigl( \omega^{\frac{1}{p}} \bigl\vert {f^{\Delta}} \bigr\vert \bigr)^{p} ( t )\Delta t \biggr)^{\frac{1}{p}} \biggr]^{k} \\ =& \biggl( \int _{0}^{h} \omega^{1 - q} \Delta t \biggr)^{\frac{k}{q}} \biggl( \int _{0}^{h} \bigl( \omega \bigl\vert f^{\Delta}\bigr\vert \bigr)^{p} ( t )\Delta t \biggr)^{\frac{k}{p}}. \end{aligned}$$
□
Theorem 2.3
Let
\(\omega ( t )\)
be positive and continuous on
\(( 0,h )\), with
\(\int _{0}^{h} \omega^{1 - q} ( t )\nabla t < \infty\), \(q > 1\). For differentiable
\(f: [ 0,h ] \to\mathbb{R}\)
with
\(f ( 0 ) = 0\)
we have
$$ \int _{0}^{h} \bigl\vert {f^{k} } \bigr\vert ^{\nabla}\nabla t \le \biggl( \int _{0}^{h} \omega^{1 - q} \nabla t \biggr)^{\frac{k}{q}} \biggl( \int _{0}^{h} \omega \bigl\vert {f^{\nabla}} \bigr\vert ^{p} \nabla t \biggr)^{\frac{k}{q}}, $$
(4)
where
\(p > 1\)
and
\(\frac{1}{p} + \frac{1}{q} = 1\).
Theorem 2.4
Assume that
\(p > 1\), \(q = \frac{p}{p - 1}\), \(\alpha \in [ 0,1 ]\), \(h \in ( 0,\infty )_{\mathbb{T}}\), \(\omega \in\mathbb{C} ( [ 0,h ]_{\mathbb{T}} , ( 0,\infty ) )\)
and
\(f \in \mathbb{C}_{\diamondsuit_{\alpha}}^{1} ( [ 0,h ]_{\mathbb{T}} ,\mathbb{R} )\). If
\(\alpha f^{\Delta}\ge 0\), \(( 1 - \alpha )f^{\nabla}\ge0\)
and
\(f ( 0 ) = 0\)
then
$$\begin{aligned}& \alpha^{k} \int _{0}^{h} \bigl\vert { \bigl( f^{k} \bigr)^{\Delta}( t )} \bigr\vert \Delta t + ( 1 - \alpha )^{k} \int _{0}^{h} \bigl\vert { \bigl( f^{k} \bigr)^{\nabla}( t )} \bigr\vert \nabla t \\& \quad \le \biggl( \int _{0}^{h} \omega^{1 - q} ( t ) \diamondsuit _{\alpha}t \biggr)^{\frac{k}{q}} \biggl( \int _{0}^{h} \omega( t ) \bigl\vert f^{\diamondsuit _{\alpha}} ( t ) \bigr\vert ^{p} \diamondsuit_{\alpha}t \biggr)^{\frac{k}{p}}. \end{aligned}$$
(5)
Proof
By Theorems 2.2, 2.3, Hölder’s inequality and \(k = \frac{k}{q} + ( 1 + p )\frac{k}{p}\), we get
$$\begin{aligned}& \alpha^{k} \int _{0}^{h} \bigl\vert \bigl( f^{k} \bigr)^{\Delta}( t ) \bigr\vert \Delta t + ( 1 - \alpha )^{k} \int _{0}^{h} \bigl\vert \bigl( f^{k} \bigr)^{\nabla}( t ) \bigr\vert \nabla t \\& \quad = \alpha^{\frac{k}{q} + ( 1 + p )\frac{k}{p}} \int _{0}^{h} \bigl\vert \bigl( f^{k} \bigr)^{\Delta}( t ) \bigr\vert \Delta t + ( 1 - \alpha )^{\frac{k}{q} + ( 1 + p )\frac{k}{p}} \int _{0}^{h} \bigl\vert \bigl( f^{k} \bigr)^{\nabla}( t ) \bigr\vert \nabla t \\& \quad \le\alpha^{\frac{k}{q} + ( 1 + p )\frac{k}{p}} \biggl( \int _{0}^{h} \omega^{1 - q} ( t )\Delta t \biggr)^{\frac{k}{q}} \biggl( \int _{0}^{h} \omega ( t ) \bigl\vert f^{\Delta}( t ) \bigr\vert ^{p} \Delta t \biggr)^{\frac{k}{p}} \\& \qquad{}+ ( 1 - \alpha )^{\frac{k}{q} + ( 1 + p )\frac{k}{p}} \biggl( \int _{0}^{h} \omega^{1 - q} ( t )\nabla t \biggr)^{\frac{k}{q}} \biggl( \int _{0}^{h} \omega ( t ) \bigl\vert f^{\nabla}( t ) \bigr\vert ^{p} \nabla t \biggr)^{\frac{k}{p}} \\& \quad \le \biggl( \alpha \int _{0}^{h} \omega^{1 - q} ( t )\Delta t \biggr)^{\frac{k}{q}} \biggl( \alpha \int _{0}^{h} \omega ( t ) \bigl\vert \alpha f^{\Delta}( t ) + ( 1 - \alpha )f^{\nabla}( t ) \bigr\vert ^{p} \Delta t \biggr)^{\frac{k}{p}} \\& \qquad{}+ \biggl( ( 1 - \alpha ) \int _{0}^{h} \omega^{1 - q} ( t )\nabla t \biggr)^{\frac{k}{q}} \\& \qquad{}\cdot \biggl( ( 1 - \alpha ) \int _{0}^{h} \omega ( t ) \bigl\vert \alpha f^{\Delta}( t ) + ( 1 - \alpha )f^{\nabla}( t ) \bigr\vert ^{p} \nabla t \biggr)^{\frac{k}{p}} \\& \quad = \biggl( \alpha \int _{0}^{h} \omega^{1 - q} ( t )\Delta t \biggr)^{\frac{k}{q}} \biggl( \alpha \int _{0}^{h} \omega ( t ) \bigl\vert f^{\diamondsuit _{\alpha}} ( t ) \bigr\vert ^{p} \Delta t \biggr)^{\frac{k}{p}} \\& \qquad{}+ \biggl( ( 1 - \alpha ) \int _{0}^{h} \omega^{1 - q} ( t )\nabla t \biggr)^{\frac{k}{q}} \biggl( ( 1 - \alpha ) \int _{0}^{h} \omega ( t ) \bigl\vert f^{\diamondsuit_{\alpha}} ( t ) \bigr\vert ^{p} \nabla t \biggr)^{\frac{k}{p}} \\& \quad \le \biggl[ \biggl( \alpha \int _{0}^{h} \omega^{1 - q} ( t )\Delta t \biggr)^{k} + \biggl( ( 1 - \alpha ) \int _{0}^{h} \omega^{1 - q} ( t )\nabla t \biggr)^{k} \biggr]^{\frac{1}{q}} \\& \qquad{}\cdot \biggl[ \biggl( \alpha \int _{0}^{h} \omega ( t ) \bigl\vert f^{\diamondsuit_{\alpha}} ( t ) \bigr\vert ^{p} \Delta t \biggr)^{k} + \biggl( ( 1 - \alpha ) \int _{0}^{h} \omega ( t ) \bigl\vert f^{\diamondsuit_{\alpha}} ( t ) \bigr\vert ^{p} \nabla t \biggr)^{k} \biggr]^{\frac{1}{q}} \\& \quad \le \biggl( \alpha \int _{0}^{h} \omega^{1 - q} ( t )\Delta t + ( 1 - \alpha ) \int _{0}^{h} \omega^{1 - q} ( t )\nabla t \biggr)^{\frac{k}{q}} \\& \qquad{}\cdot \biggl( \alpha \int _{0}^{h} \omega ( t ) \bigl\vert f^{\diamondsuit_{\alpha}} ( t ) \bigr\vert ^{p} \Delta t + ( 1 - \alpha ) \int _{0}^{h} \omega ( t ) \bigl\vert f^{\diamondsuit_{\alpha}} ( t ) \bigr\vert ^{p} \nabla t \biggr)^{\frac{k}{p}} \\& \quad = \biggl( \int _{0}^{h} \omega^{1 - q} ( t ) \diamondsuit_{\alpha}t \biggr)^{\frac{k}{q}} \biggl( \int _{0}^{h} \omega ( t ) \bigl\vert f^{\diamondsuit _{\alpha}} ( t ) \bigr\vert ^{p} \diamondsuit_{\alpha}t \biggr)^{\frac{k}{p}} . \end{aligned}$$
□
Theorem 2.5
Assume that
\(1 < p \le2\), \(q = \frac{p}{p - 1}\), \(\alpha \in [ 0,1 ]\), \(h \in ( 0,\infty )_{\mathbb{T}}\), \(\omega \in\mathbb{C} ( [ 0,h ]_{\mathbb{T}} , ( 0,\infty ) )\)
and
\(f \in\mathbb{C}_{\diamondsuit_{\alpha}}^{1} ( [ 0,h ]_{\mathbb{T}} ,\mathbb{R} )\). If
\(\alpha f^{\Delta}\ge 0\), \(( 1 - \alpha )f^{\nabla}\ge0\)
and
\(f ( 0 ) = 0\), then
$$ \begin{aligned}[b] &\alpha^{k} \int _{0}^{u} \bigl\vert { \bigl( f^{k} \bigr)^{\Delta}( t )} \bigr\vert \Delta t + ( 1 - \alpha )^{k} \int _{0}^{u} \bigl\vert { \bigl( f^{k} \bigr)^{\nabla}( t )} \bigr\vert \nabla t \\ &\quad\le\sum_{j = 0}^{k - 2} \alpha^{j} \left ( \begin{matrix} k \\ j \end{matrix} \right ) \gamma^{j} \beta ^{\frac{k - j}{q}} \biggl[ \int _{0}^{h} \omega( t ) \bigl\vert g^{\diamondsuit_{\alpha}} ( t ) \bigr\vert ^{p} \diamondsuit_{\alpha}t \biggr] ^{\frac{k - 1}{p}} \\ &\qquad{}+ \alpha^{k - 1} \left ( \begin{matrix} k \\ k - 1 \end{matrix} \right )\gamma^{k - 1} \bigl( f ( h ) - f ( 0 ) \bigr), \end{aligned} $$
(6)
where
\(\beta: = \min_{u \in [ 0,h ]_{T} } v ( u )\), \(v ( u ) = \max \{ \int _{0}^{u} \omega^{1 - q} ( t )\diamondsuit _{\alpha}t , \int _{u}^{h} \omega^{1 - q} ( t )\diamondsuit_{\alpha}t \} \), \(\gamma: = \max \{ \vert f ( 0 ) \vert , \vert f ( h ) \vert \} \).
Proof
We let \(u \in [ 0,h ]_{\mathbb{T}}\) be arbitrary. By applying Theorem 2.4 to the function \(g ( t ) = f ( t ) - f ( 0 )\), we obtain
$$\begin{aligned}& \alpha^{k} \int _{0}^{u} \bigl\vert \bigl( f^{k} \bigr)^{\Delta}( t ) \bigr\vert \Delta t + ( 1 - \alpha )^{k} \int _{0}^{u} \bigl\vert \bigl( f^{k} \bigr)^{\nabla}( t ) \bigr\vert \nabla t \\& \quad = \alpha^{k} \int _{0}^{u} \left \vert \sum _{j = 0}^{k - 1} \left ( \begin{matrix} k \\ j \end{matrix} \right ) \bigl( g^{k - j} \bigr)^{\Delta}f^{j} ( 0 ) \right \vert \Delta t \\& \qquad{}+ ( 1 - \alpha )^{k} \int _{0}^{u} \left \vert \sum _{j = 0}^{k - 1} \left ( \begin{matrix} k \\ j \end{matrix} \right ) \bigl( g^{k - j} \bigr)^{\nabla}f^{j} ( 0 ) \right \vert \nabla t \\& \quad \le \left ( \begin{matrix} k \\ 0 \end{matrix} \right ) \biggl[ \alpha^{k} \int _{0}^{u} \bigl\vert g^{k} \bigr\vert ^{\Delta}\Delta t + ( 1 - \alpha )^{k} \int _{0}^{u} \bigl\vert g^{k} \bigr\vert ^{\nabla}\nabla t \biggr] \\& \qquad{}+ \alpha \left ( \begin{matrix} k \\ 1 \end{matrix} \right ) \bigl\vert f ( 0 ) \bigr\vert \biggl[ \alpha^{k - 1} \int _{0}^{u} \bigl\vert g^{k - 1} \bigr\vert ^{\Delta}\Delta t + ( 1 - \alpha )^{k - 1} \int _{0}^{u} \bigl\vert g^{k - 1} \bigr\vert ^{\nabla}\nabla t \biggr] \\& \qquad{}+ \alpha^{2} \left ( \begin{matrix} k \\ 2 \end{matrix} \right ) \bigl\vert f^{2} ( 0 ) \bigr\vert \biggl[ \alpha^{k - 2} \int _{0}^{u} \bigl\vert g^{k - 2} \bigr\vert ^{\Delta}\Delta t + ( 1 - \alpha )^{k - 2} \int _{0}^{u} \bigl\vert g^{k - 2} \bigr\vert ^{\nabla}\nabla t \biggr] \\& \qquad{}\vdots \\& \qquad{}+ \alpha^{k - 2} \left ( \begin{matrix} k \\ k - 2 \end{matrix} \right ) \bigl\vert f^{k - 2} ( 0 ) \bigr\vert \biggl[ \alpha^{2} \int _{0}^{u} \bigl\vert g^{2} \bigr\vert ^{\Delta}\Delta t + ( 1 - \alpha )^{2} \int _{0}^{u} \bigl\vert g^{2} \bigr\vert ^{\nabla}\nabla t \biggr] \\& \qquad{}+ \alpha^{k - 1} \left ( \begin{matrix} k \\ k - 1 \end{matrix} \right ) \bigl\vert f^{k - 1} ( 0 ) \bigr\vert \biggl[ \alpha \int _{0}^{u} \vert f \vert ^{\Delta}\Delta t + ( 1 - \alpha ) \int _{0}^{u} \vert f \vert ^{\nabla}\nabla t \biggr] \\& \quad \le \left ( \begin{matrix} k \\ 0 \end{matrix} \right ) \biggl[ \int _{0}^{u} \omega^{1 - q} ( t ) \diamondsuit_{\alpha}t \biggr]^{\frac{k}{q}} \biggl[ \int _{0}^{u} \omega ( t ) \bigl\vert g^{\diamondsuit_{\alpha}} ( t ) \bigr\vert ^{p} \diamondsuit_{\alpha}t \biggr]^{\frac{k}{p}} \\& \qquad{}+ \alpha \left ( \begin{matrix} k \\ 1 \end{matrix} \right ) \bigl\vert f ( 0 ) \bigr\vert \biggl[ \int _{0}^{u} \omega^{1 - q} ( t ) \diamondsuit_{\alpha}t \biggr]^{\frac{k - 1}{q}} \biggl[ \int _{0}^{u} \omega ( t ) \bigl\vert g^{\diamondsuit_{\alpha}} ( t ) \bigr\vert ^{p} \diamondsuit_{\alpha}t \biggr]^{\frac{k - 1}{p}} \\& \qquad{} + \alpha^{2} \left ( \begin{matrix} k \\ 2 \end{matrix} \right ) \bigl\vert f^{2} ( 0 ) \bigr\vert \biggl[ \int _{0}^{u} \omega^{1 - q} ( t ) \diamondsuit_{\alpha}t \biggr]^{\frac{k - 2}{q}} \biggl[ \int _{0}^{u} \omega ( t ) \bigl\vert g^{\diamondsuit_{\alpha}} ( t ) \bigr\vert ^{p} \diamondsuit_{\alpha}t \biggr]^{\frac{k - 2}{p}} \\& \qquad{}\vdots \\& \qquad{} + \alpha^{k - 2} \left ( \begin{matrix} k \\ k - 2 \end{matrix} \right ) \bigl\vert f^{k - 2} ( 0 ) \bigr\vert \biggl[ \int _{0}^{u} \omega^{1 - q} ( t ) \diamondsuit_{\alpha}t \biggr]^{\frac{2}{q}} \\& \qquad{}\cdot \biggl[ \int _{0}^{u} \omega ( t ) \bigl\vert g^{\diamondsuit_{\alpha}} ( t ) \bigr\vert ^{p} \diamondsuit_{\alpha}t \biggr]^{\frac{2}{p}} \\& \qquad{}+ \alpha^{k - 1} \left ( \begin{matrix} k \\ k - 1 \end{matrix} \right ) \bigl\vert f^{k - 1} ( 0 ) \bigr\vert \alpha \int _{0}^{u} \bigl\vert f^{\Delta}( t ) \bigr\vert \Delta t \\& \qquad{}+ \alpha^{k - 1} \left ( \begin{matrix} k \\ k - 1 \end{matrix} \right ) \bigl\vert f^{k - 1} ( 0 ) \bigr\vert ( 1 - \alpha ) \int _{0}^{u} \bigl\vert f^{\nabla}( t ) \bigr\vert \nabla t \\& \quad \le\sum_{j = 0}^{k - 2} \alpha^{j} \left ( \begin{matrix} k \\ j \end{matrix} \right ) \bigl\vert f^{j} ( 0 ) \bigr\vert \biggl[ \int _{0}^{u} \omega^{1 - q} ( t ) \diamondsuit_{\alpha}t \biggr]^{\frac{k - j}{q}} \biggl[ \int _{0}^{u} \omega ( t ) \bigl\vert g^{\diamondsuit_{\alpha}} ( t ) \bigr\vert ^{p} \diamondsuit_{\alpha}t \biggr]^{\frac{k - j}{p}} \\& \qquad{} + \alpha^{k - 1} \left ( \begin{matrix} k \\ k - 1 \end{matrix} \right )\gamma^{k - 1} \bigl( f ( u ) - f ( 0 ) \bigr). \end{aligned}$$
Similarly,
$$\begin{aligned}& \alpha^{k} \int _{u}^{h} \bigl\vert \bigl( f^{k} \bigr)^{\Delta}( t ) \bigr\vert \Delta t + ( 1 - \alpha )^{k} \int _{u}^{h} \bigl\vert \bigl( f^{k} \bigr)^{\nabla}( t ) \bigr\vert \nabla t \\& \quad \le\sum_{j = 0}^{k - 2} \alpha^{j} \left ( \begin{matrix} k \\ j \end{matrix} \right ) \bigl\vert f^{j} ( 0 ) \bigr\vert \biggl[ \int _{u}^{h} \omega^{1 - q} ( t ) \diamondsuit _{\alpha}t \biggr]^{\frac{k - j}{q}} \biggl[ \int _{u}^{h} \omega ( t ) \bigl\vert g^{\diamondsuit_{\alpha}} ( t ) \bigr\vert ^{p} \diamondsuit_{\alpha}t \biggr]^{\frac{k - j}{p}} \\& \qquad{}+ \alpha^{k - 1} \left ( \begin{matrix} k \\ k - 1 \end{matrix} \right )\gamma^{k - 1} \bigl( f ( h ) - f ( u ) \bigr). \end{aligned}$$
Adding these two inequalities and taking into account that \(a^{r} + b^{r} \le ( a + b )^{r}\) holds, for \(a,b \ge0\) and \(r \ge 1\), yield the desired inequality. □